AP Calculus AB Free Response Notebook Table of Contents Area and Volume ................................................................. 1-25 Charts with Riemann Sums, MVT, Ave. Rates/Values ........... 41-53 Analyzing Graph of f’ ............................................................ 54-59 Slope Fields with differential Equations ............................... 60-70 Related Rates ....................................................................... 71-77 Accumulation Functions....................................................... 78-91 Implicit Differentiation ......................................................... 92-97 Particle Motion .................................................................... 98-108 Charts of f, f’, f’’ ................................................................... 109-113 Functions/Misc. ................................................................... 114-128 1998 AP Calculus AB Scoring Guidelines 1. Let R be the region bounded by the x{axis, the graph of y = px, and the line x = 4. (a) Find the area of the region R. (b) Find the value of h such that the vertical line x = h divides the region R into two regions of equal area. (c) Find the volume of the solid generated when R is revolved about the x{axis. (d) The vertical line x = k divides the region R into two regions such that when these two regions are revolved about the x{axis, they generate solids with equal volumes. Find the value of k. (a) y 3 4 y = px 8 1: A = Z px dx 2 > 0 2 < 1: answer :> 1 R O 1 2 3 4 5 x 4 4 2 3=2 16 A = Z px dx = x = or 5:333 3 3 0 0 h 8 h 4 (b) Z px dx = Z px dx = Z px dx 1: equation in h 0 3 0 h 2 ( {or{ 1: answer 2 8 2 16 2 h3=2 = h3=2 = h3=2 3 3 3 3 − 3 h = p3 16 or 2:520 or 2:519 4 4 x2 2 (c) V = π Z (px) dx = π = 8π 1: limits and constant 0 2 8 0 > 3 <> 1: integrand or 25.133 or 25.132 > 1: answer :> k k 4 2 2 2 (d) π Z (px) dx = 4π π Z (px) dx = π Z (px) dx 1: equation in k 0 0 k 2 ( {or{ 1: answer k2 k2 k2 π = 4π π = 8π π 2 2 − 2 k = p8 or 2:828 Copyright ©1998 College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 1 AB{2 / BC{2 1999 2 The shaded region, R , is b ounded by the graph of y = x and the line 2. y y x 2 y =4,asshown in the gure ab ove. (a) Find the area of R . y 4 (b) Find the volume of the solid generated by revolving R ab out the x{axis. There exists a number k , k>4, such that when R is revolved ab out (c) x y = k , the resulting solid has the same volume as the solid in the line O part (b). Write, but do not solve, an equation involving an integral expression that can b e used to nd the value of k . Z 2 ( 2 1: integral (4 x ) dx (a) Area = 2 2 1: answer Z 2 2 = 2 (4 x ) dx 0 2 3 x = 2 4x 3 0 32 = =10:666 or 10:667 3 Z 2 8 2 2 2 1: limits and constant 4 (x ) dx (b) Volume = > > < 2 1: integrand 3 Z > 2 > : 4 1: answer = 2 (16 x ) dx 0 2 5 x = 2 16x 5 0 256 = = 160:849 or 160:850 5 Z 2 8 256 2 2 2 1: limits and constant dx = (c) (k x ) (k 4) > > > 5 2 > < 2: integrand 4 < 1 > each error > > > > : 1: equation 2 !0Å#ALCULUSÅ!"nÅÅ"#n y ,ETÅ2ÅBEÅTHEÅSHADEDÅREGIONÅINÅTHEÅFIRSTÅQUADRANTÅENCLOSEDÅBYÅTHEÅGRAPHSÅOF YE X ÅYXCOS ÅANDÅTHEÅY AXIS ÅASÅSHOWNÅINÅTHEÅFIGUREÅABOVE 1.5 A &INDÅTHEÅAREAÅOFÅTHEÅREGIONÅ2 1 2 ye= – x B &INDÅTHEÅVOLUMEÅOFÅTHEÅSOLIDÅGENERATEDÅWHENÅTHEÅREGIONÅ2ÅISÅREVOLVED ABOUTÅTHEÅX AXIS 0.5 R yx= 1 – cos C 4HEÅREGIONÅ2ÅISÅTHEÅBASEÅOFÅAÅSOLIDÅ&ORÅTHISÅSOLID ÅEACHÅCROSSÅSECTION x O 0.5 1 1.5 PERPENDICULARÅTOÅTHEÅX AXISÅISÅAÅSQUAREÅ&INDÅTHEÅVOLUMEÅOFÅTHISÅSOLID 2EGIONÅ2 Å #ORRECTÅLIMITSÅINÅANÅINTEGRALÅINÅA ÅB EXX COSÅATÅXÅÅÅÅ! ORÅC ! X ¦£ ÅINTEGRAND A !REA ¨ COSEXDX Ť¦ ¦ ÅANSWER ¥¦ ÅÅORÅ ! £ ÅINTEGRANDÅANDÅCONSTANT B 6OLUME Å Q ¨ EXDXX COS ¦ ¦ Ť¦ÅÅÅÅÅ ÅEACHÅERROR ¦ ¦ ÅANSWER Å Q ÅÅÅORÅ ¥¦ ! ¦£ ÅINTEGRAND X ¦ C 6OLUME ¨ EXDX COS ¦ ¦ÅÅÅÅÅ ÅEACHÅERROR ¦ ¦ÅÅÅÅÅÅ.OTEÅÅIFÅNOTÅOFÅTHEÅFORM ŠŤ¦ ¦ ¦ D ¦ÅÅÅÅÅÅÅÅÅÅÅÅÅKFXGXDX¨ ¦ C ¦ ÅANSWER ¥¦ Copyright © 2000 by College Entrance Examination Board and Educational Testing Service. All rights reserved. AP is a registered trademark of the College3 Entrance Examination Board. AP® CALCULUS AB 2001 SCORING GUIDELINES Question 1 Let R and S be the regions in the first quadrant shown in the figure above. The region R is bounded by the x-axis and the graphs of yx2 3 and yx tan . The region S is bounded by the y-axis and the graphs of yx2 3 and yx tan . (a) Find the area of R. (b) Find the area of S. (c) Find the volume of the solid generated when S is revolved about the x-axis. Point of intersection 2tanxx3 at (AB , ) (0.902155,1.265751) A 3 2 ¦£ 1 : limits (a) Area R = ¨¨tanxdx 2 x3 dx = 0.729 ¦ 0 A ¦ 3 : ¤ 1 : integrand or ¦ ¦ 1 : answer B ¥¦ Area R = ¨ (2yydy )1/3 tan 1 = 0.729 0 or 3 2 A Area R = ¨¨22tanxdx33 x xdx = 0.729 00 A £ 3 ¦ 1 : limits (b) Area S = ¨ 2tanxxdx = 1.160 or 1.161 ¦ 0 ¦ 3 : ¤ 1 : integrand or ¦ ¦ B 2 ¥¦ 1 : answer Area S = ¨¨tan11/3ydy (2 y ) dy = 1.160 or 1.161 0 B or Area S 2 B = ¨¨(2ydy )1/3 (2 y ) 1/3 tan 1 ydy 00 = 1.160 or 1.161 A £ 322 ¦ 1 : limits and constant (c) Volume = 3¨ 2tanxxdx ¦ 0 ¦ 3 : ¤ 1 : integrand = 2.6523 or 8.331 or 8.332 ¦ ¦ ¥¦ 1 : answer Copyright © 2001 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 4 2 AP® CALCULUS AB 2002 SCORING GUIDELINES (Form B) Question 1 Let 4 be the region bounded by the O-axis and the graphs of N ! O and ON" , as shown in the figure above. N (a) Find the area of 4. (b) Find the volume of the solid generated when 4 is revolved about the N-axis. (c) The region 4 is the base of a solid. For this solid, each cross section perpendicular to the N-axis is a square. Find the volume of this solid. Region 4 1 : Correct limits in an integral in (a), (b), or (c). N ! " N at N = 1.487664 = ) N ) ¬! £ 1 : integrand N ­ ¦ (a) Area = ¨ " N@N­ 2 ¤¦ ® N ­ ¦ 1 : answer ¥¦ = 3.214 or 3.215 (b) Volume ¦£ 2 : integrand and constant ) ¬¬N ! ¦ = 3 (4 2N@N ) ­ ­ ¦ ¨ ­ ­ ¦ ® ®1 N ­ ­ 3 ¤ < 1 > each error ¦ ¦ 1 : answer = 31.884 or 31.885 or 10.1493 ¥¦ ) ¬N ! ¦£ 2 : integrand (c) Volume = " N@N­ ¦ ¨ ­ ¦ ® N ­ ¦ 1 each error ¦ ¦ note: 0/2 if not of the form = 8.997 3 ¤¦ ¦ ¦ @ ¦ ¦ kfxgxdx¨ ( ( ) ( )) ¦ ? ¦ 1 : answer ¥¦ Copyright © 2002 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 5 2 AP® CALCULUS AB 2002 SCORING GUIDELINES Question 1 Let B and C be the functions given by BN AN and CN( )ln. N (a) Find the area of the region enclosed by the graphs of B and C between N and N 1. (b) Find the volume of the solid generated when the region enclosed by the graphs of B and C between N and N is revolved about the line O 4. (c) Let D be the function given by DN( ) BN( ) CN( ). Find the absolute minimum value of DN on the 1 closed interval >>N 1, and find the absolute maximum value of DN on the closed interval 2 1 >>N 1. Show the analysis that leads to your answers. 2 £ 1 : integral (a) Area = AN@NN ln = 1.222 or 1.223 ¦ ¨ 2 ¤ ¦ 1 : answer ¥¦ ¦£ 1 : limits and constant (b) Volume = 3 4lnNA@N 4 N ¦ ¨ ¦ ¦ 2 : integrand ¦ = 7.5153 or 23.609 ¦ 1 each error ¦ 4 ¤¦ ¦ Note: 0 / 2 if not of the form ¦ ¦ > ¦ kRxrxdx( ) ( ) ¦ ¨= ¦ ¦ 1 : answer ¥¦ £ 1: considers DNa( ) 0 (c) DNaaa B N CN AN ¦ N ¦ ¦ 1 : identifies critical point ¦ N 0.567143 3 ¤ ¦ and endpoints as candidates ¦ ¦ 1: answers Absolute minimum value and absolute ¥¦ maximum value occur at the critical point or at the endpoints. Note: Errors in computation come off the third point. D(0.567143) 2.330 D(0.5) 2.3418 D(1) 2.718 The absolute minimum is 2.330. The absolute maximum is 2.718. Copyright © 2002 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks6 of the College Entrance Examination Board. 2 AP® CALCULUS AB 2003 SCORING GUIDELINES Question 1 Let R be the shaded region bounded by the graphs of yx= and ye= 3x and the vertical line x = 1, as shown in the figure above. (a) Find the area of R. (b) Find the volume of the solid generated when R is revolved about the horizontal line y = 1. (c) The region R is the base of a solid. For this solid, each cross section perpendicular to the x-axis is a rectangle whose height is 5 times the length of its base in region R. Find the volume of this solid. Point of intersection 1: Correct limits in an integral in ex3x = at (T, S) = (0.238734, 0.488604) (a), (b), or (c) 1 £ 1 : integrand (a) Area = ()xe 3x dx ¦ ¨T 2 : ¤ ¦ 1 : answer = 0.442 or 0.443 ¥ 1 3x 2 2 £ 2 : integrand (b) Volume = Q ()11exdx () ¦ ¨T ()¦ ¦ < 1> reversal = 0.453 Q or 1.423 or 1.424 ¦ ¦ < 1> error with constant 3 : ¤¦ ¦ < 1> omits 1 in one radius ¦ ¦ < 2> other errors ¦ ¦ 1 : answer ¥¦ (c) Length = xe 3x ¦£ 2 : integrand ¦ Height = 5()xe 3x ¦ < 1 > incorrect but has ¦ ¦ 3x 3 : ¤ xe ¦ 1 ¦ 3x 2 ¦ as a factor Volume = ¨ 5()xe dx = 1.554 ¦ T ¦ 1 : answer ¥¦ Copyright © 2003 by College Entrance Examination Board.